Problem: Solve for $x$ : $4x^2 + 12x - 160 = 0$
Solution: Dividing both sides by $4$ gives: $ x^2 + {3}x {-40} = 0 $ The coefficient on the $x$ term is $3$ and the constant term is $-40$ , so we need to find two numbers that add up to $3$ and multiply to $-40$ The two numbers $-5$ and $8$ satisfy both conditions: $ {-5} + {8} = {3} $ $ {-5} \times {8} = {-40} $ $(x {-5}) (x + {8}) = 0$ Since the following equation is true we know that one or both quantities must equal zero. $(x -5) (x + 8) = 0$ $x - 5 = 0$ or $x + 8 = 0$ Thus, $x = 5$ and $x = -8$ are the solutions.